Correct Answer - Option 4 : 45
Concept:
\(\rm sin^2\theta+cos^2\theta=1\)
Calculation:
Given:
a cos θ + b sin θ = 3 .......(1)
a sin θ - b cos θ = 6 ........(2)
Now squaring (1) and (2), we get
\(\rm a^2cos^2\theta+2ab\ sin\theta \ cos\theta +b^2sin^2\theta=9\) .... (3)
\(\rm a^2sin^2\theta-2ab\ sin\theta \ cos\theta +b^2cos^2\theta=36\) .... (4)
Adding (3) and (4), we get
\(\rm\rm [a^2cos^2\theta+2ab\ sin\theta \ cos\theta +b^2sin^2\theta]+ [a^2sin^2\theta-2ab\ sin\theta \ cos\theta +b^2cos^2\theta]=9+36\)
\(\rm a^2+b^2\) = 45 (∵ \(\rm sin^2\theta+cos^2\theta=1\))
Hence, option (4) is correct.
